Answer : Tension in the line = 936.7 N
Explanation :
It is given that,
Mass of student, m = 65 kg
The angle between slackline and horizontal, 
The two forces that acts are :
(i) Tension
(ii) Weight
So, from the figure it is clear that :




Hence, this is the required solution.
Answer:
85.31 N
Explanation:
Given,
Radius of Earth = 3960 miles = 6373 Km
Weight of Astronaut at Sea level = 200lb = 90.72 Kg
Altitude of Astronaut above Earth = 125 miles = 201.17 km
We know that,
-------------------------- (1)
where,
F = force on the object due to gravity also called the weight
m = mass of the object = 200 lb = 90.71 kg
g = acceleration due to gravity = 9.8 m/s²
Also,
-------------------(2)
where,
F = force due to gravity
G = Gravitational constant =
Nm²/kg²
M = mass of the Earth = 
r = distance between the two objects
here, r = (6373+201.17)km = 6574170 m
From equation (1),

Putting value of m in equation (2)


Answer:
A) T1 = 566 k = 293°C
B) T2 = 1132 k = 859°C
Explanation:
A)
The average kinetic energy of the molecules of an ideal gas is givwn by the formula:
K.E = (3/2)KT
where,
K.E = Average Kinetic Energy
K = Boltzman Constant
T = Absolute Temperature
At 10°C:
K.E = K10
T = 10°C + 273 = 283 K
Therefore,
K10 = (3/2)(K)(283)
FOR TWICE VALUE OF K10:
T = T1
Therefore,
2 K10 = (3/2)(K)(T1)
using the value of K10:
2(3/2)(K)(283) = (3/2)(K)(T1)
<u>T1 = 566 k = 293°C</u>
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B)
The average kinetic energy of the molecules of an ideal gas is given by the formula:
K.E = (3/2)KT
but K.E is also given by:
K.E = (1/2)(m)(vrms)²
Therefore,
(3/2)KT = (1/2)(m)(vrms)²
vrms = √(3KT/m)
where,
vrms = Root Mean Square Velocity of Molecule
K = Boltzman Constant
T = Absolute Temperature
m = mass
At
T = 10°C + 273 = 283 K
vrms = √[3K(283)/m]
FOR TWICE VALUE OF vrms:
T = T2
Therefore,
2 vrms = √(3KT2/m)
using the value of vrms:
2√[3K(283)/m] = √(3KT2/m)
2√283 = √T2
Squaring on both sides:
(4)(283) = T2
<u>T2 = 1132 k = 859°C</u>